Difference between revisions of "2005 AMC 10A Problems/Problem 10"
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== Solution 3== | == Solution 3== | ||
There is only one positive value for k such that the quadratic equation would have only one solution. | There is only one positive value for k such that the quadratic equation would have only one solution. | ||
− | k-8 and -k-8 are the values of a. I am certain we | + | k-8 and -k-8 are the values of a. I am certain we all know what -8-8 is. That’s correct, <math>\implies \boxed{A}.</math> |
==See also== | ==See also== |
Revision as of 16:06, 2 September 2019
Problem
There are two values of for which the equation has only one solution for . What is the sum of those values of ?
Solution 1
A quadratic equation has exactly one root if and only if it is a perfect square. So set
Two polynomials are equal only if their coefficients are equal, so we must have
or .
So the desired sum is
Alternatively, note that whatever the two values of are, they must lead to equations of the form and . So the two choices of must make and so and .
Solution 2
Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the sign when added). So we must have Therefore, we have \implies \boxed{A}.$
Solution 3
There is only one positive value for k such that the quadratic equation would have only one solution. k-8 and -k-8 are the values of a. I am certain we all know what -8-8 is. That’s correct,
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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